Solve the second equation for I_1 and the third equation for I_3. Then plug those into the first equation. From here you can solve for I_2. Then you can plug that in to equation 2 to solve for I_1. Then plug in to equation 3 to solve for I_3.
a) 6.52 m/s^2
b) 23.47 km/hr^2
a) v = v0 + at --> 9 = 0 + a(1.38) --> a = 6.52
b) 6.52 m/s2 * 60 s2 / 1 min2 * 60 min2 / 1 hr2 = 23.47 km/hr2
Answer:
The phase difference between these two waves is 141.1⁰
Explanation:
The displacement of the wave is given as;

Amplitude, A = 2yₓCos(¹/₂Φ)
Since the amplitude of the combination is 1.5 times that of one of the original amplitudes = yₓ = 1.5 × A = 1.5A
A = 2(1.5A)Cos(¹/₂Φ)
A = 3ACos(¹/₂Φ)
¹/₃ = Cos(¹/₂Φ)
(¹/₂Φ) = Cos ⁻(0.3333)
(¹/₂Φ) = 70.55°
Φ = 141.1°
The phase difference between these two waves is 141.1⁰
Increase as density increase and vise versa.
<span>The wavelength increases when a sound wave travels from a less dense to a more dense medium, the speed increases, and the frequency stays the same.</span>